In a previous question of mine, I asked how the volume of the region $$xy>zw \quad\wedge \quad x>-y$$ bounded by the unit $4$-ball (so $x^2+y^2+z^2+w^2<1$) could be calculated.
The answer to my question implied that the answer ($1/4$ the volume of the unit $4$-ball) could easily be seen through symmetry, but I never really understood the arguments, hence this question.
I'd really appreciate either a new perspective on this, or an explanation of the answerers argument, which roughly goes as follows:
- The condition $0<x+y$ divides the circle (the area of the projection of the total region on the $x,y$ plane, given the values of $z$ and $w$) in half (on the $x,y$ plane).
- Folding the negative $z$ half plane onto the positive one, to get the complementary superposition, divides the volume in half (on the $z,w$ plane).
The total result is to divide the whole ball by $4$.
I believe I understand the first part, but the second part eludes me; I cannot seem to visualize this "folding".
I think the idea is to say that for every pair of values of $(z,w)$ and $(-z,w),$ the sum of the areas of the projections of the region in question onto the $x,y$ plane for these two pairs ($(z,w)$ and $(-z,w)$) is a quarter of the area of the circle, or something like that. The answerer writes "the sum of the two areas, for the same $|z \cdot w|$ value, totals that of the semicircle." But I can't see why this should be true from the symmetries of the region.